Problem 19 - Calculus

Question

Suppose that \(f(x)\) is a function with \(f(100)=35\) and \(f^{\prime}(100)=3 .\) Estimate \(f(102)\).

Step-by-Step Solution

Answer

\( f(102) \approx 41 \).

Step 1: Understand the Problem

We are given that \( f(x) \) is a function, and we know its value and derivative at \( x = 100 \). We need to use these to estimate \( f(102) \).

Step 2: Recall the Linear Approximation Formula

The linear approximation formula for a function \( f \) at a point \( x = a \) is given by: \( f(x) \approx f(a) + f'(a)(x - a) \). This formula provides an estimate of the function near the point \( a \).

Step 3: Apply the Formula

Since we know \( f(100) = 35 \) and \( f'(100) = 3 \), we can use the linear approximation formula to estimate \( f(102) \):\[f(102) \approx f(100) + f'(100)(102 - 100)\]Substitute the given values:\[f(102) \approx 35 + 3 \times 2\]

Step 4: Compute the Estimation

Calculate the expression from the previous step:\[f(102) \approx 35 + 6 = 41\]Thus, \( f(102) \) is approximately \( 41 \).

Key Concepts

Function Derivatives

Understanding function derivatives is key to many calculus concepts, including linear approximation. A derivative of a function, denoted as \( f'(x) \), represents the rate at which the function is changing at any given point. Think of it as the "slope" or "instantaneous rate of change" of the function at a certain point.

For example, in the exercise, we are given that \( f'(100) = 3 \). This means at \( x = 100 \), the function \( f(x) \) is increasing at a rate of 3 units for every single unit increase in \( x \).

When you have a function and its derivative, you can use this information to approximate values of the function near the given point by considering how the function is changing. This leads us to the concept of linear approximation, which relies on these derivatives.

Function Evaluation

Function evaluation is a fundamental step in calculus, where you determine the output of a function given a specific input. In simpler terms, you plug a number into a function to see what number comes out.

With the exercise example, we have the function \( f(x) \) and we know \( f(100) = 35 \). This evaluation tells us that when \( x \) is 100, the function outputs 35.

By combining these evaluations with derivatives, you can use them in more complex calculations like linear approximation. Evaluation provides you the fixed starting point – knowing the exact value of the function at a particular point helps establish the groundwork for further estimation processes.

Estimation Techniques

Estimation techniques in mathematics allow us to work with approximate values when exact values are difficult to calculate. Linear approximation is one such technique that helps us estimate function values using simpler linear relationships. It’s particularly useful when fetching exact calculations might be too complicated or time-consuming.

This method uses a linear equation derived from the known value and derivative of the function at a particular point. The formula \( f(x) \approx f(a) + f'(a)(x - a) \) essentially gives us a line that is tangent to the function at \( x = a \). This tangent line is used to estimate the function value at points near \( a \).

In the given exercise, this approach allows us to estimate \( f(102) \) using the known values at \( x = 100 \). By substituting \( f(100) = 35 \) and \( f'(100) = 3 \) into the formula, we quickly find that \( f(102) \approx 41 \). This kind of estimation is incredibly helpful in both practical and theoretical applications where simplification and speed are needed.